50W enough? pretty much in the ballpark. I'm a regular guy myself, but what's 6dB between friends?
a.wayne will love this discussion (if you haven't already gone through that, haven't read the last few pages)

And it had what I love to hear most in any amp, it somehow suggested that it had infinite power, as if you simply couldn't overload it anyhow. An illusion, of course, but quite a number of American amps pull this off, and I daresay nobody pulls it off like the Yanks. Well, maybe Naim a little bit.

That's exactly what all systems should sound like: I found it amazing, 20 years ago, trying some monster amp that would crush my toes if it fell on them, start to collapse as you pushed the volume up. A "miserable", 20W, chip amp should give you that impression if it has a decent power supply; it's what's called "engineering" ...

As I understood it from an admittedly brief lookover (no time now, will dive in later on), you got 53,050 uF per channel for a nominally 100W/8 Ohms amp?

Tom, as I see it, no amp ever made satisfied this criteria, or perhaps a few wildcard products did, but I never heard about them.

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Hi dvv,

The calculation example that you quoted was not calculating the required reservoir capacitance for a power supply. It was merely to help illustrate what types of calculations could be performed, with an equation that had been derived.

In that particular example, the result meant that IF you needed a capacitance to supply all of the current for the first zero-to-peak rise of a 5-Amp-Peak 30 Hz sine wave, then if that capacitance was not at least 53050 uF, the voltage across the capacitance would drop by at least one volt.

The result was subsequently verified with an LT-Spice simulation.

That equation was actually just an approximate result; a sort of "worst-case estimation" equation:

(13b): C ≥ a / (πfΔv_MAX)

where a is the 0-to-peak amplitude of the capacitor's sinusoidal current waveform, f is frequency in Hertz, and C is capacitance in Farads.

The more-complete picture, for the scenario being considered in that derivation, is this one:

(17): C ≥ Δi / ( 2πf∙(Δv - (ESR∙Δi)))

Equation (17) gives the capacitance value, C, that would be required in order to supply the current for the first quarter-cycle of a sine signal of frequency f (in Hz), with 0-to-peak amplitude Δi Amperes, while causing the voltage across the capacitor to dip by no more than your choice of Δv Volts.

To use equation (17), it will be easier to first set ESR to zero, calculate a C value, find an estimate for ESR for that C value at the frequency being used, and then re-calculate the C value with the ESR value.

If we still want to just account for the whole sine wave, we should be able to simply double the C value given by equation (17), since we're considering only the positive or negative half-cycle, but not both, and the other half of the half-cycle of a sine wave is symmetrical and thus encloses the same area (its integral), i.e. the same amp-seconds value, as the first half.

[Edit: Note, too, that positive capacitor current was defined as current flowing into the positive-voltage-designated lead of the capacitor. So, typically, for this scenario, both Δi and Δv will be negative.]

For more of this exciting story, including some of the considerations involved when Δv - (ESR∙Δi) gets close to zero, go to